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kvasek [131]
3 years ago
5

When a liquid is at its boiling point there is an increase in potential energy, kinetic energy, or all of the above

Physics
1 answer:
Darya [45]3 years ago
6 0

Answer:potential energy

Explanation:

When a liquid is at it boiling point, there is increase in it potential energy

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rodikova [14]
The force of gravity between two objects is increased when the mass of objects is increased.
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4 years ago
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archaeologists use radioactive decay of an isotope of the element because it has a fairly short half-life and is found in anythi
natulia [17]

Archaeologists use radioactive decay of an isotope of the element <u>carbon.</u> Because it has a fairly short half-life, and is found in anything that was once living.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

Archaeologists use radioactive decay of an isotope of the element <u>carbon.</u> Because it has a fairly short half-life, and is found in anything that was once living.

Hence, <u>carbon</u> is used the correct answer for the blank.

To learn more about the radioactivity, refer  to the link;

brainly.com/question/1770619

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7 0
2 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
3 years ago
For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red
barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

z=\sqrt{\frac{c+v}{c-v}}-1

- A red dwarf moving away from Earth at 39.1 km/s :

z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}

- A blue dwarf moving away from Earth at 25.9 km/sz=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}

- A red dwarf moving transversely at 14.1 km/s

z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}

6 0
4 years ago
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ruslelena [56]
The answer to your question is SMOOTH
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4 years ago
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